$$ \sin x+\sin y=a \text { and } \cos x+\cos y=b \text {. Prove that } \tan ^{2}\left(\frac{x-y}{2}\right)=\frac{4-a^{2}-b^{2}}{a^{2}+b^{2}} \text {. } $$

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written 2.1 years ago by

Solution:

Given, $ a=\sin x+\sin y b=\cos x+\cos y\\ $

$$ \begin{aligned} &\therefore \mathrm{a}^{2}+\mathrm{b}^{2} \\\\ &=[\sin \mathrm{x}+\sin \mathrm{y}]^{2}+[\cos \mathrm{x}+\cos \mathrm{y}]^{2} \\\\ &\text { Formula } \mathrm{S}+\mathrm{S}=2 \mathrm{SC} \quad \mathrm{C}+\mathrm{C}=2 \mathrm{CC} \\\\ &\left.=\left[2 \sin \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\right]^{2}+2 \cos \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos \left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)\right]^{2} \\\\ &=4 \sin ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right)+4 \cos ^{2}\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right) \cos ^{2}\left(\frac{\mathrm{x}-\mathrm{y}}{2}\right) …

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